Question 19 of 239

The recoil speed of a hydrogen atom after it emits a photon in going from $n=5$ state to $n=5$ state will be

A$4.17m{s}^{-1}$B$2.19m{s}^{-1}$C$4.34m{s}^{-1}$D$3.25m{s}^{-1}$

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Answer: (A) $4.17m{s}^{-1}$

$\left(\Delta E\right)$ Releases when photon going from $n=5$ to $n=\Delta E$
$=\left(13.6-0.54\right)eV=13.06eV$

${P}_{i}={P}_{f}$ (By linear momentum conservation)
$0=\dfrac{h}{\lambda }-Mv={V}_{Recoil}=\dfrac{h}{\lambda M}...\left(1\right)$
&\Delta E=\dfrac{hc}{\lambda }=\dfrac{hc}{\lambda M}\times M\Rightarrow Mc{V}_{Recoil}
${V}_{Recoil}=\dfrac{\Delta E}{Mc}=\dfrac{13.06\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}\times 3\times {10}^{8}}=4.17m{s}^{-1}$

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